# resistor and inductor in series

#### Gokee2

Joined Jun 5, 2009
7
I have a 267 Ohm resistor and 10 mH inductor in series. I am feeding the end of the resistor with output from a soundcard. All grounds are after the inductor. I am taking readings though my line in port and looking at them with xoscope. I have channel 1 plugged into the same spot as the function generator at the beginning of the resistor and channel two at the other end of the resistor at the inductor.​
I take my readings on a sine wave with a frequency of 1k. Channel1 is at 1.24 V and channel 2 at 280 mV.​
Get the phase shift with inverse cosine (C2 / C1) (C for Channel) and get 76.9 Degree's.​
Next I get Inductive Reactance with Resistor * tan(Phase shift) and get 1.15K​
Then I get the Inductance with (Inductive Reactance)/(2(pie)frequency) and end up with 183 mH!​
Why dont I end up with something close to the 10 mH I put in?​
Thanks!​

#### mik3

Joined Feb 4, 2008
4,843
Can you make it more clear how you calculated the inductance?

#### Gokee2

Joined Jun 5, 2009
7
phase shift
inverse cosine (C2 / C1)​
cos-1(.28/1.24)=​
cos-1(.2258)=76.95 This is a phase shift of 76.95 degrees

Inductive Reactance​
Resistor * tan(Phase shift)​
267tan(76.95)=1.15k​

Inductance​
(inductive Reactance)/(2(pie)frequency)​
1150/(2(pie)1000) = 183 mH

That better?

#### Gokee2

Joined Jun 5, 2009
7
Why does this place allow .doc files but not .odt?

Ok this is the lab I am working on. I decided to go ahead and put in more numbers today. My question is why is the inductance changing according to my numbers? Did I do some calculation wrong?

Thanks!

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#### mik3

Joined Feb 4, 2008
4,843
I have never used your method to make the calculations.

However, if you use the voltage divider rule you will find that the inductance is 10mH.

0.28=(1.24*XL)/(XL+270)

Solve for XL.

#### JoeJester

Joined Apr 26, 2005
4,390
Why does this place allow .doc files but not .odt?
.odt's are templates, from open office, there probably isn't a need to allow a template.

#### Gokee2

Joined Jun 5, 2009
7
I have never used your method to make the calculations.

However, if you use the voltage divider rule you will find that the inductance is 10mH.

0.28=(1.24*XL)/(XL+270)

Solve for XL.

Sorry its been awhile since algebra and I was never very good at it. I am not sure how to get XL out by itself.

Also if I do how would I get from XL to L? Or was that part of mine right? I had never heard of the "voltage divider rule" before now.

Thanks!

#### JoeJester

Joined Apr 26, 2005
4,390
Gokee2,

I looked at your table 1 Figure 1 information.

I know it doesn't matter, but your instructions stated you were setting the generator to 2 Vpk, but that is immaterial for now.

Using only the 1 kHz frequency, your output agrees closely with what you should have gotten. I say that because you never completed the measurements you read when you measured each component.

Your value for Xl is incorrect because you used the X' phase angle vice the R phase angle when you did the calculation. That of course threw your XL and L out the window.

Your first clue with the calculations should have been the measured value of the coil before you started the lab.

#### Gokee2

Joined Jun 5, 2009
7
Gokee2,

I looked at your table 1 Figure 1 information.

I know it doesn't matter, but your instructions stated you were setting the generator to 2 Vpk, but that is immaterial for now.
I know, however we had talked about that in class and he said to bring it as high as you can with the sine wave still looking good. At just above 1.28 V the sine wave starts clipping, no matter how high I turn up my soundcard it does not get much higher and it looks a lot like a square wave. Kinda a problem of using a soundcard for a oscilloscope. The lab was written for the day classes where students use a real digital oscilloscope.

Using only the 1 kHz frequency, your output agrees closely with what you should have gotten. I say that because you never completed the measurements you read when you measured each component.

Your value for Xl is incorrect because you used the X' phase angle vice the R phase angle when you did the calculation. That of course threw your XL and L out the window.
Ahha! That makes sense! I did 90-76.9 and used that and ended up with 9.87 mH!

Your first clue with the calculations should have been the measured value of the coil before you started the lab.
Ya, I should have got the lab done sooner and measured the value of the coil at school.

Thanks for your help in figuring this out. I think I understand it now. Too bad I had to turn it in as it was a few hours ago.

Last edited:

#### Gokee2

Joined Jun 5, 2009
7
.odt's are templates, from open office, there probably isn't a need to allow a template.
Sorry got to correct you on this one. .odt is Open Document Text. Templetes are .ott (at least in openofficeI am not sure if the templates are part of the ODF standard or not). See http://en.wikipedia.org/wiki/OpenDocument

odt files can come from many places including:

openoffice
MS office with the odf plugin http://www.sun.com/software/star/odf_plugin/
Abiword
koffice
many others

#### Ron H

Joined Apr 14, 2005
7,014
I have never used your method to make the calculations.

However, if you use the voltage divider rule you will find that the inductance is 10mH.

0.28=(1.24*XL)/(XL+270)

Solve for XL.
You need to remember that this is a vector calculation. The simple voltage divider equation ignores the fact that the phase of the voltage across the inductor is not the same as the input voltage. It will not yield the correct value for $$X_L$$.
You can solve for $$X_L$$ using this equation:
$$\frac{Vout}{Vin}=\frac{X_L}{\sqrt{R^2+X_L^2}}$$